M.C.Q (1 Marks)

MCQ 11 Mark

If $f(x) = ax + b$, where a and b are integers, $f(–1) = –5$ and $f(3) = 3,$ then $a$ and $b$ are equal to.

A
$a = –3, b = –1$
✓
$a = 2, b = –3$
C
$a = 0, b = 2$
D
$a = 2, b = 3$

Answer

Correct option: B.

$a = 2, b = –3$

Given that: $f(x) = ax + b$
$\Rightarrow f(-1) = a(-1) + b$
$\Rightarrow -5 = -a + b$
$\Rightarrow a - b = 5 ...........(i)$
$f(3) = 3a + b$
$\Rightarrow 3 = 3a + b$
$\Rightarrow 3a + b = 3 ........(ii)$
On solving eqn. $(i)$ and $(ii),$ We get $a = 2, b = -3$

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MCQ 21 Mark

The domain and range of real function $f$ defined by $\text{f(x)}=\sqrt{\text{x}-1}$ is given by.

✓
Domain $= [1, \infty),$ Range $= [0, \infty)$
B
Domain $= [1, \infty),$ Range $= [0, \infty)$
C
Domain $= [1, \infty),$ Range $= [0, \infty)$
D
Domain $= [1, \infty),$ Range $= [0, \infty)$

Answer

Correct option: A.

Domain $= [1, \infty),$ Range $= [0, \infty)$

We have, $\text{f(x)}=\sqrt{\text{x}-1}$
Clearly, $f(x)$ is defined if $\text{x}-1\geq0$
$\Rightarrow\text{x}\geq1$
$\therefore$ Domain of $\text{f}=[1, \infty)$
Now for $\text{x}\geq1,\text{x}-1\geq0$
$\Rightarrow\sqrt{\text{x}-1}\geq1$
$\Rightarrow$ Range of $= [0, \infty)$

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MCQ 31 Mark

The domain of the function $f$ defined by $\text{f(x)}=\sqrt{4-\text{x}}+\frac{1}{\sqrt{\text{x}^2-1}}$ is equal to.

✓
$(–\infty, –1) \cup (1, 4]$
B
$(–\infty, –1] \cup (1, 4]$
C
$(–\infty, –1) \cup [1, 4]$
D
$(–\infty, –1) \cup [1, 4)$

Answer

Correct option: A.

$(–\infty, –1) \cup (1, 4]$

We have, $\text{f(x)}=\sqrt{4-\text{x}}+\frac{1}{\sqrt{\text{x}^2-1}}$
$f(x)$ is defined if $4 - \text{x}\geq 0$ and $\text{x}^2-1>0$
$\Rightarrow\text{x}\leq4$ and $(\text{x}+1)(\text{x}-1)>0$
$\Rightarrow\text{x}\leq4$ and $(\text{x}<-1 \ \text{or} \ \text{x}>1)$
$\therefore$ Domain of $\text{f}=(-\infty, -1)\cup(1, 4]$

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MCQ 41 Mark

Domain of $\sqrt{\text{a}^2-\text{x}^2}(\text{a}>0)$ is.

A
$(-a, a)$
✓
$[-a, a]$
C
$[0, a]$
D
$(-a, 0]$

Answer

Correct option: B.

$[-a, a]$

We have $\text{f(x)}\sqrt{\text{a}^2-\text{x}^2}$
Clearly $f(x)$ is defined, if ${\text{a}^2-\text{x}^2}\geq0$
$\Rightarrow\text{x}^2\leq\text{a}^2$
$\Rightarrow-\text{a}\leq\text{x}\leq\text{a} \ [\therefore\text{a}>0]$
$\therefore$ Domain of $f$ is $[-a, a]$

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MCQ 51 Mark

The domain of the function $f$ given by $\text{f(x)}=\frac{\text{x}^2+2\text{x}+1}{\text{x}^2-\text{x}-6}.$

✓
$R – \{3, –2\}$
B
$R – \{–3, 2\}$
C
$R – [3, –2]$
D
$R – (3, –2)$

Answer

Correct option: A.

$R – \{3, –2\}$

Given that: $\text{f(x)}=\frac{\text{x}^2+2\text{x}+1}{\text{x}^2-\text{x}-6}$
$f(x)$ is defined if $\text{x}^2-\text{x}-6\neq0$
$\Rightarrow\text{x}^2-3\text{x}+2\text{x}-6\neq0$
$\Rightarrow(\text{x}-3)(\text{x}+2)\neq0$
$\Rightarrow\text{x}\neq-2,\text{x}\neq3$
So, the domain of $f(x) = R - \{-2, 3\}$

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MCQ 61 Mark

If $[x]^2 - 5[x] + 6 = 0,$ where $[.]$ denote the greatest integer function$,$ then.

A
$x \in [3, 4]$
B
$x \in (2, 3]$
✓
$x \in [2, 3]$
D
$x \in [2, 4]$

Answer

Correct option: C.

$x \in [2, 3]$

We have $[x]^2 - 5[x] + 6 = 0$
$\Rightarrow [x]^2 - 3[x] 2[x] + 6 = 0$
$\Rightarrow [x]([x] - 3) -2([x] - 3) = 0$
$\Rightarrow ([x] - 3)([x] - 2) = 0$
$\Rightarrow [x] = 2, 3$
So$, x \in [2, 3]$

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MCQ 71 Mark

Let $n(A) = m,$ and $n(B) = n.$ Then the total number of non$-$empty relations that can be defined from $A$ to $B$ is.

A
$m^n$
B
$n^m - 1$
C
$mn - 1$
✓
$2^{mn} - 1$

Answer

Correct option: D.

$2^{mn} - 1$

We have$, n(A) = mn$ and $n(B) = n$
$n(A \times B) = n(A).n(B) = mn$
Total number of relation from $A$ to $B =$ Number of subsets of$ A \times B = 2^{mn}$
So$,$ total number if non$-$empty relations $= 2^{mn} - 1$

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MCQ 81 Mark

The domain and range of the real function $f$ defined by $\text{f(x)}=\frac{4-\text{x}}{\text{x}-4}$ is given by.

A
Domain $= R,$ Range $= \{-1, 1\}$
B
Domain $= R - \{1\},$ Range $= R$
✓
Domain $= R - \{4\},$ Range $= \{-1\}$
D
Domain $= R - \{-4\}$, Range $= \{-1, 1\}$

Answer

Correct option: C.

Domain $= R - \{4\},$ Range $= \{-1\}$

Given that: $\text{f(x)}=\frac{4-\text{x}}{\text{x}-4}$
We know that $f(x)$ is defined if $\text{x}-4\neq0$
$\Rightarrow \text{x}\neq4$
So, the domain of $f(x)$ is $= R - \{4\}$
Let $\text{f(x)}=\text{y}=\frac{4-\text{x}}{\text{x}-4}$
$\Rightarrow\text{yx}-4\text{y}=4-\text{x}$
$\Rightarrow\text{yx}+\text{x}=4\text{y}+4$
$\Rightarrow\text{x}(\text{y}+1)=4\text{y}+4$
$\Rightarrow\text{x}=\frac{4(1+\text{y})}{1+\text{y}}$
If $x$ is real number, then $1+\text{y}\neq0$
$\Rightarrow\text{x}\neq1$
$\therefore$ Range of $f(x) = R - \{-1\}$

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MCQ 91 Mark

Range of $\text{f(x)}=\frac{1}{1-2\cos\text{x}}$ is.

A
$\Big[\frac{1}{3}, 1\Big]$
B
$\Big[-1, \frac{1}{3}\Big]$
✓
$(-\infty, -1]\cup\Big[\frac{1}{3},\infty\Big)$
D
$\Big[-\frac{1}{3}, 1\Big]$

Answer

Correct option: C.

$(-\infty, -1]\cup\Big[\frac{1}{3},\infty\Big)$

We know that, $-1\leq-\cos\text{x}\leq1$
$\Rightarrow-1\leq-\cos\text{x}\leq1$
$\Rightarrow-2\leq-2\cos\text{x}\leq2$
$\Rightarrow-1\leq-2\cos\text{x}\leq3$
Now $\text{f(x)}=\frac{1}{1-2\cos\text{x}}$ is defined if
$-1\leq-2\cos\text{x}\leq0$ or $0<1-2\cos\text{x}\leq3$
$\Rightarrow-1\geq\frac{1}{1-2\cos\text{x}}>-\infty$ or $\infty>\frac{1}{1-2\cos\text{x}}\geq\frac{1}{3}$
$\Rightarrow\frac{1}{1-2\cos\text{x}}\in(-\infty, -1]\cup\Big[\frac{1}{3},\infty\Big)$

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MCQ 101 Mark

The domain for which the functions defined by $ f(x) = 3x^2 – 1$ and $g(x) = 3 + x$ are equal is.

A
$\Big\{-1, \frac{4}{3}\Big\}$
✓
$\Big[-1, \frac{4}{3}\Big]$
C
$\Big(-1, -\frac{4}{3}\Big)$
D
$\Big[-1, -\frac{4}{3}\Big]$

Answer

Correct option: B.

$\Big[-1, \frac{4}{3}\Big]$

We have$, f(x) = 3x^2 – 1$ and $g(x) = 3 + x$
$f(x) = g(x)$
$\Rightarrow 3x^2 – 1 = 3 + x$
$\Rightarrow 3x^2 – x - 4 = 0$
$\Rightarrow (3x - 4)(x + 1) = 0$
$\therefore\text{x}=-1, \frac{4}{3}$

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MCQ 111 Mark

Let $\text{f(x)}=\sqrt{1+\text{x}^2}$ then.

A
$\text{f(xy)} = \text{f(x)}.\text{f(y)}$
B
$\text{f(xy)} \geq \text{f(x)}.\text{f(y)}$
✓
$\text{f(xy)} \leq \text{f(x)}.\text{f(y)}$
D
None of these

Answer

Correct option: C.

$\text{f(xy)} \leq \text{f(x)}.\text{f(y)}$

Given that: $\text{f(x)}=\sqrt{1+\text{x}^2}$
$\Rightarrow\text{f(xy)}=\sqrt{1+\text{x}^2\text{y}^2}$
and $\text{f(x)}.\text{f(y)}=\sqrt{1+\text{x}^2}.\sqrt{1+\text{x}^2}=\sqrt{1+\text{x}^2 +\text{y}^2+\text{x}^2\text{y}^2}$
$\therefore\sqrt{1+\text{x}^2\text{y}^2}\leq\sqrt{1+\text{x}^2 +\text{y}^2+\text{x}^2\text{y}^2}$
$\text{f(xy)} \leq \text{f(x)}.\text{f(y)}$

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MCQ 121 Mark

The domain and range of the function $f$ given by $f(x) = 2 - x - 5|$ is.

A
Domain $= R^+, $Range $= ( –\infty, 1)$
✓
Domain $= R,$ Range $= ( –\infty, 2)$
C
Domain $= R,$ Range $= ( –\infty, -2)$
D
Domain$ = R^{+ },$ Range $= ( –\infty, 2)$

Answer

Correct option: B.

Domain $= R,$ Range $= ( –\infty, 2)$

We have$, f(x) = 2 - |x - 5|$
Clearly$, f(x)$ is defined for all $\text{x}\in\text{R}.$
$\therefore$ Domain of $f = R$
Now$, |\text{x}-5|\geq0,\forall\text{x}\in\text{R}$
$\Rightarrow-|\text{x}-5|\leq0$
$\Rightarrow2-|\text{x}-5|\leq2$
$\therefore\text{f(x)}\leq2$
$\therefore$ Range of $\text{f}=(-\infty, 2)$

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Maths M.C.Q (1 Marks)

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